mat25-hw6solns

mat25-hw6solns - Math 25: Advanced Calculus UC Davis,...

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Math 25: Advanced Calculus UC Davis, Spring 2011 Math 25 Solutions to Homework Assignment #6 1. Let sequences ( s n ) n =1 and ( t n ) n =1 converge to respective limits S and T . Fix some > 0. By definition, there exist numbers m s ,m t such that for any n s m s and n t m t , | s n s- S | < 2 and | t n t- T | < 2 . If n max { m s ,m t } , by the triangle inequality, | ( s n- t n )- ( S- T ) | = | ( s n- S )- ( t n- T ) | | s n- S | + | t n- T | < 2 + 2 = . By definition, then, s n- t n- S- T as n- . 2. For ( s n ) n =1 and ( t n ) n =1 such that s n = n + 1 , t n = n, Both sequences diverge to infinity, and hence the reasoning in problem 1. does not apply. However, for any n , since a- b = ( a- b )( a + b ), s n- t n = n + 1- n = ( n + 1)- n n + 1 + n = 1 n + 1 + n Now, since 1 n + 1 + n 1 n And because 1 n- 0 as n- , by the squeeze theorem, ( s n- t n ) n...
View Full Document

This note was uploaded on 10/16/2011 for the course MATH 34 taught by Professor Wiley during the Winter '11 term at UC Merced.

Page1 / 2

mat25-hw6solns - Math 25: Advanced Calculus UC Davis,...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online